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A five-person team is to be formed from a pool of 6 East All Stars and

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A five-person team is to be formed from a pool of 6 East All Stars and  [#permalink]

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New post 06 Apr 2016, 07:07
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A
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C
D
E

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  55% (hard)

Question Stats:

65% (02:37) correct 35% (02:58) wrong based on 83 sessions

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A five-person team is to be formed from a pool of 6 East All Stars and  [#permalink]

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New post 06 Apr 2016, 08:00
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A five-person team is to be formed from a pool of 6 East All Stars and 6 West All Stars. What is the probability that the team will contain at least 2 East All Stars?

The question with at- least can be easily tackled by
At least = Total cases - unwanted cases

So atleast 2 player = total cases - cases with no east all stars person - case with only one east all person

Total number of cases = 12C5 = 798

Cases with no east all person = 6C5(All person from west all stars) = 6
Cases with exactly one east all person = 6C4*6C1 (4from west all stars and 1 from east all stars)= 90

So total no. of cases with with atleast two members = 798-6-90 = 702

Probability = 702/798 = 29/33

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Re: A five-person team is to be formed from a pool of 6 East All Stars and  [#permalink]

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New post 06 Apr 2016, 08:33
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Teams can be formed as 2E-3W , 3E-2W , 4E-W and 5E
Possible ways to make a team 2E-3W = 6c2 * 6c3 = 300
Possible ways to make a team 3E-2W = 6c3 * 6c2 = 300
Possible ways to make a team 4E-W = 6c4 * 6c1 = 90
Possible ways to make a team 5E = 6c5 = 6

Number of ways to select a team of 5 out of 12 people = 12c5
=12!/(7!*5!)
= 792

Probability that the team will contain at least 2 East All Stars
= (300+300+90+6) / 792
=696/792
= 29/33

Answer A

Alternatively we can also calculate the number of possible ways to make a team of 5W and E-4W .
Possible ways to make a team 5W = 6c5 = 6
Possible ways to make a team E-4W = 6c1 * 6c4 = 90

Probability that the team will contain at least 2 East All Stars = 1 - (96/792)
= 29/33
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A five-person team is to be formed from a pool of 6 East All Stars and  [#permalink]

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New post 18 Mar 2017, 00:48
Bunuel wrote:
A five-person team is to be formed from a pool of 6 East All Stars and 6 West All Stars. What is the probability that the team will contain at least 2 East All Stars?

A. 29/33
B. 81/169
C. 57/120
D. 47/144
E. 119/720


Hi Bunuel, Math Experts,

Ive been trying to find a pattern to understand when do we need to multiply the counts by the number of arrangements and when not to.
For example, in this question, while counting the probability of selecting a team that contains 3 east all stars and 2 west all stars, what I did was :

\(\frac{6C3*6C3}{12C3}\) * \(\frac{5!}{3!*2!}\)

Similarly, I counted the numbers for a team with 2 east all stars, 4 east all stars and all 5 east all stars.

Can you please help understand the concept ?

For ex., why did we multiply by the no. of arrangements in qsn : https://gmatclub.com/forum/there-are-8- ... 92342.html but not in the current one ?
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Re: A five-person team is to be formed from a pool of 6 East All Stars and  [#permalink]

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New post 10 Sep 2017, 12:10
Bunuel wrote:
A five-person team is to be formed from a pool of 6 East All Stars and 6 West All Stars. What is the probability that the team will contain at least 2 East All Stars?

A. 29/33
B. 81/169
C. 57/120
D. 47/144
E. 119/720


It is sometimes really good thing to check options fist.
For ex in this case we could save us a lot of time.

All these probabilities except that one in A are less than 1/2. It just can't be. So the answer is A
Why? How many players from East can be in our command? 0, 1, 2, 3, 4, or 5.
Which ones fit in our case? 2, 3, 4, 5- it is definetely more than and 1.
(We know that 6C6 = 6C0, 6C5 = 6C1 and so on).
So the probaility will be decently greater than 1/2.
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Re: A five-person team is to be formed from a pool of 6 East All Stars and   [#permalink] 10 Sep 2017, 12:10
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